Question

The work function of a metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The maximum velocity of emitted photoelectron will be……..

(A) 10 ms^–1

(B) 10³ms^–1

(C) 10⁴ms^–1

(D) 10⁶ms^–1

Answer 1

The correct option is (D) 10⁶ms^–1.

Explanation:

 ɸ = 1 eV = 1.6 × 10^–19 J

λ = 3000 Å = 3000 × 10^–10 m

we know (1/2)mV_max² = (hc / λ) – ɸ

= [{6.62 × 10^–34 × 3 × 10⁸} / {3000 × 10^–10}] – (1.6 × 10^–19)

∴ (1/2) mV_max² = {(6.62 × 10^–26) / (10^–7)} – 1.6 × 10^–19

= 6.62 × 10^–19 – 1.6 × 10^–19

= 5.02 × 10^–19 J

∴ V_max = √[(2 × 5.02 × 10^–19) / (9.1 × 10^–31)]

= 1.05 × 10⁶ m/s

= 10⁶ m/s